3.503 \(\int \frac{1}{x^2 (a+b x^2)^{3/2}} \, dx\)

Optimal. Leaf size=38 \[ -\frac{2 b x}{a^2 \sqrt{a+b x^2}}-\frac{1}{a x \sqrt{a+b x^2}} \]

[Out]

-(1/(a*x*Sqrt[a + b*x^2])) - (2*b*x)/(a^2*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.0079569, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {271, 191} \[ -\frac{2 b x}{a^2 \sqrt{a+b x^2}}-\frac{1}{a x \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^2)^(3/2)),x]

[Out]

-(1/(a*x*Sqrt[a + b*x^2])) - (2*b*x)/(a^2*Sqrt[a + b*x^2])

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{1}{x^2 \left (a+b x^2\right )^{3/2}} \, dx &=-\frac{1}{a x \sqrt{a+b x^2}}-\frac{(2 b) \int \frac{1}{\left (a+b x^2\right )^{3/2}} \, dx}{a}\\ &=-\frac{1}{a x \sqrt{a+b x^2}}-\frac{2 b x}{a^2 \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0055127, size = 27, normalized size = 0.71 \[ -\frac{a+2 b x^2}{a^2 x \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x^2)^(3/2)),x]

[Out]

-((a + 2*b*x^2)/(a^2*x*Sqrt[a + b*x^2]))

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Maple [A]  time = 0.003, size = 26, normalized size = 0.7 \begin{align*} -{\frac{2\,b{x}^{2}+a}{{a}^{2}x}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b*x^2+a)^(3/2),x)

[Out]

-(2*b*x^2+a)/x/(b*x^2+a)^(1/2)/a^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.27512, size = 70, normalized size = 1.84 \begin{align*} -\frac{{\left (2 \, b x^{2} + a\right )} \sqrt{b x^{2} + a}}{a^{2} b x^{3} + a^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

-(2*b*x^2 + a)*sqrt(b*x^2 + a)/(a^2*b*x^3 + a^3*x)

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Sympy [A]  time = 0.788244, size = 46, normalized size = 1.21 \begin{align*} - \frac{1}{a \sqrt{b} x^{2} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{2 \sqrt{b}}{a^{2} \sqrt{\frac{a}{b x^{2}} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(b*x**2+a)**(3/2),x)

[Out]

-1/(a*sqrt(b)*x**2*sqrt(a/(b*x**2) + 1)) - 2*sqrt(b)/(a**2*sqrt(a/(b*x**2) + 1))

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Giac [A]  time = 2.61432, size = 68, normalized size = 1.79 \begin{align*} -\frac{b x}{\sqrt{b x^{2} + a} a^{2}} + \frac{2 \, \sqrt{b}}{{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

-b*x/(sqrt(b*x^2 + a)*a^2) + 2*sqrt(b)/(((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)*a)